How far should an object be kept from the concave mirror of focal length 20 cm to form a real image one-fifth of its size?

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Solution

$m = (- \frac{1}{5}) = - \frac{V}{U}$

$\Rightarrow \frac{V}{U} = \frac{1}{5}$

$\Rightarrow V = \frac{U}{5}$

Now by applyng the mirror formula
$\frac{1}{U} + \frac{1}{V} = \frac{1}{f}$

$\frac{1}{U} + \frac{5}{U} = \frac{- 1}{20}$

$\frac{6}{U} = \frac{- 1}{20}$
$U = - 120 c m$

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How far should an object be kept from the concave mirror of focal length 20 cm to form a real image one-fifth of its size?

m=(−15)=−VU⇒VU=15⇒V=U5Now by applyng the mirror formula1U+1V=1f1U+5U=−1206U=−120U=−120cm

$m = (- \frac{1}{5}) = - \frac{V}{U}$

$\Rightarrow \frac{V}{U} = \frac{1}{5}$

$\Rightarrow V = \frac{U}{5}$

Now by applyng the mirror formula
$\frac{1}{U} + \frac{1}{V} = \frac{1}{f}$

$\frac{1}{U} + \frac{5}{U} = \frac{- 1}{20}$

$\frac{6}{U} = \frac{- 1}{20}$
$U = - 120 c m$